Basically since there’s 2 hydrogen’s there will be two H’s on either side of your other element. And Se is in group 6 which means it has 6 valence electrons. When you combine 6 and 2 from the hydrogen you get 8. You then should place 8 dots around Se two on each side.
So something like H Se(with 8 dogs around it) H
Answer:
A. 3.2L of NO
B. 4.8L of H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
4NH3 + 5O2 —> 4NO + 6H2O
A. Determination of the litres of NO produced from the reaction. This is illustrated below:
From the balanced equation above,
5L of O2 produced 4L of NO.
Therefore, 4L of O2 will produce = (4x4)/5 = 3.2L of NO.
B. Determination of the litres of H2O produced from the reaction. This is illustrated below:
From the balanced equation above,
5L of O2 produced 6L of H2O.
Therefore, 4L of O2 will produce = (4x6)/5 = 4.8L of H2O
Answer:
![[A_t]=54.5\ g](https://tex.z-dn.net/?f=%5BA_t%5D%3D54.5%5C%20g)
Explanation:
Given that:
Half life = 14.0 days
Where, k is rate constant
So,
The rate constant, k = 0.04951 days⁻¹
Initial concentration [A₀] = 60.0 g
Time = 46.7 hrs
Considering, 1 hr = 0.041667 days
So, time = 1.9458 days
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
So,
![[A_t]=60.0\times e^{-0.04951\times 1.9458}\ g](https://tex.z-dn.net/?f=%5BA_t%5D%3D60.0%5Ctimes%20e%5E%7B-0.04951%5Ctimes%201.9458%7D%5C%20g)
Noble gases are known for having a full outer shell of electrons which helium has as it has two electrons its first electron shell is completely filled
Yes, it can happen.
If all the lower energy orbits are already filled with electrons, then it cannot happen since the lower orbitals will already be full, hence another electron will have to move.
However, if the lower energy orbits are not full (for instance if one excited electron goes back to its normal state), then it can happen.
